The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form $$ E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2). $$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable.

There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(X;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$

However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module.

I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form $$ E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2). $$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable.

There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(SZ;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$

However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module.

I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{Z}/2$ coefficients this takes the form $$ E_2^{p,q} = H^p(BG; H^q(X;\mathbb{Z}/2))\Rightarrow H^*_G(X;\mathbb{Z}/2). $$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives the cohomology of $G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{Z}/2)$). Hence the $E_2$-term is computable.

There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(X;\mathbb{Z}/2))\to H^{p+2}(BG;H^0(X;\mathbb{Z}/2)).$$

However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{Z}/2)\to H^*_G(SZ;\mathbb{Z}/2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{Z}/2$-module.

I believe it also gives that $H^*_G(SZ;\mathbb{Z}/2)$ is free as a $H^*_G$-module.

The standard way to compute equivariant cohomology of a $G$-space $X$ is to use the spectral sequence of the fibration $$X\to EG\times_G X\to BG,$$ where the projection is induced by $X\to \ast$. With $\mathbb{F}_2$ coefficients this takes the form $$ E_2^{p,q} = H^p(BG; H^q(X;\mathbb{F}_2))\Rightarrow H^*_G(X;\mathbb{F}_2). $$ In your case, the space $X=SZ$ is connected and one-dimensional, so there are only two non-trivial rows: $q=0$ (which just gives $H^*(BG;\mathbb{F}_2) = H^*_G$) and $q=1$ (the cohomology of $G=\mathbb{Z}/2\times\mathbb{Z}/2$ with coefficients in the module $H^1(SZ;\mathbb{F}_2)$). Hence the $E_2$-term is computable.

There is only one possible non-trivial differential $$d_2:H^p(BG;H^1(X;\mathbb{F}_2))\to H^{p+2}(BG;H^0(SZ;\mathbb{F}_2))=H^{p+2}(BG;\mathbb{F}_2).$$

However, the fact that $SZ$ has $G$-fixed points (the suspension points) implies that the above fibration admits a section, and so the induced map $H^*(BG;\mathbb{F}_2)\to H^*_G(SZ;\mathbb{F}_2)$ is injective. This agrees with the edge homomorphism of the spectral sequence, implying that $d_2$ is in fact trivial. Hence $E_2=E_\infty$. There are no extension problems, since we are using field coefficients, so this gives the equivairiant cohomology as a $\mathbb{F}_2$-module.

I believe it also gives that $H^*_G(SZ;\mathbb{F}_2)$ is free as an $H^*_G$-module.